Problem: s involving definite integrals (algebraic) AP.CALC: CHA‑4 (EU), CHA‑4.D (LO), CHA‑4.D.1 (EK), CHA‑4.D.2 (EK), CHA‑4.E (LO), CHA‑4.E.1 (EK) Google Classroom Facebook Twitter Email You might need: Calculator Problem Last week, the thickness of the ice on a lake increased by $\dfrac{25}{t+3}$ centimeters per day (where $t$ is the number of days since the ice first formed). By how many centimeters did the thickness of the ice increase between $t=0$ and $t=4$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1{,}000}{441}$ (Choice B) B $\dfrac{100}{21}$ (Choice C) C $25\ln\left(\dfrac{7}{3}\right)$ (Choice D) D $25\ln\left(4\right)$
Explanation: Letting $y(t)$ be the thickness of the ice on day $t$, we are given that $y'(t)=\dfrac{25}{t+3}$. We want to find $y(4)-y(0)$. According to the Fundamental Theorem of Calculus, $\begin{aligned} y(4)-y(0)&=\int_{0}^{4} y'(t)\,dt \\\\ &=\int_{0}^{4}\left(\dfrac{25}{t+3}\right)dt \end{aligned}$ $\int_{0}^{4}\left(\dfrac{25}{t+3}\right)dt=25\ln\left(\dfrac{7}{3}\right)$ In conclusion, between $t=0$ and $t=4$ the thickness of the ice increased by $25\ln\left(\dfrac{7}{3}\right)$ centimeters.